Question

The sum of three numbers in an Arithmetic Progression is 45 and their product is 3000. What are the three numbers?

Answer 1

First number = 10, second number =15, third number =20

Solution:

Let us assume the three numbers are a, a+d, a+2d.

Now as given the sum is 45.

So, $a+a+d+a+2d = 45$

$3a+3d =45$

$3(a+d) =45$

$(a+d) =15$

Again the product is = 3000

So, $a\times(a+d)\times(a+2d) = 3000$

$\Rightarrow a\times15\times(a+2d) = 3000$

$\Rightarrow a\times(a+2d) = 200$

$\Rightarrow a\times(a+d+d) = 200$

$\Rightarrow a\times(15+d) =200$

$\Rightarrow 15a +ad =200$

$\Rightarrow 15a + a\times(15-a) =200$

$\Rightarrow +15a-a^2= 200$

$\Rightarrow a^2-30a+200 =0$

$\Rightarrow (a-20)(a-10) =0$

$\Rightarrow a = 20 \ or \ 10$

a cannot be 20 as d will be negative.

So a = 10, d = 5

Answer 2

$Let \: (a-d), a \: and \: (a+d) \: are \: three\\consecutive \: terms \: in \: A.P$

/* According to the problem given */

$Sum \: of \: the \: three\:terms = 45$

$\implies (a-d) + a + (a+d) = 45$

$\implies 3a = 45$

$\implies a = \frac{45}{3}$

$\implies a = 15\: ---(1)$

$And \\ Product \: of \: the \: terms = 3000$

$\implies (a-d)\times a \times (a+d) = 3000$

$\implies a(a^{2} - d^{2} ) = 3000$

$\implies 15( 15^{2} - d^{2}) = 3000 \: [From \: (1) ]$

$\implies 225 - d^{2} = \frac{3000}{15}$

$\implies 225 - d^{2} = 200$

$\implies 225 - 200 = d^{2}$

$\implies d^{2} = 25$

$\implies d = ± 5$

Therefore.,

Case 1:

If a = 15, d = 5

Required 3 terms of A.p :

(a-d), a , (a+d)

i.e , (15-5), 15, (15+5)

10, 15 and 20

Case 2:

If a = 15, d = -5

Required 3 terms of A.P :

20, 15 and 10

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